17N.2.AHL.TZ0.H_8

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

EITHER

$x^2=2\sec \theta$

$2x\frac{\text{d}x}{\text{d}\theta }=2\sec \theta \tan \theta$     M1A1

$\int \frac{\text{d}x}{x\sqrt{x^4-4}}$

$=\int \frac{\sec \theta \tan \theta \text{d}\theta }{2\sec \theta \sqrt{4{\sec }^2\theta -4}}$     M1A1

OR

$x=\sqrt{2}(\sec \theta {)}^{\frac{1}{2}}\left(=\sqrt{2}{(\cos \theta )}^{-\frac{1}{2}}\right)$

$\frac{\text{d}x}{\text{d}\theta }=\frac{\sqrt{2}}{2}(\sec \theta {)}^{\frac{1}{2}}\tan \theta \left(=\frac{\sqrt{2}}{2}{(\cos \theta )}^{-\frac{3}{2}}\sin \theta \right)$     M1A1

$\int \frac{\text{d}x}{x\sqrt{x^4-4}}$

$=\int \frac{\sqrt{2}{(\sec \theta )}^{\frac{1}{2}}\tan \theta \text{d}\theta }{2\sqrt{2}{(\sec \theta )}^{\frac{1}{2}}\sqrt{4{\sec }^2\theta -4}}\left(=\int \frac{\sqrt{2}{(\cos \theta )}^{-\frac{3}{2}}\sin \theta \text{d}\theta }{2\sqrt{2}{(\cos \theta )}^{-\frac{1}{2}}\sqrt{4{\sec }^2\theta -4}}\right)$     M1A1

THEN

$=\frac{1}{2}\int \frac{\tan \theta \text{d}\theta }{2\tan \theta }$     (M1)

$=\frac{1}{4}\int \text{d}\theta$

$=\frac{\theta }{4}+c$     A1

$x^2=2\sec \theta \Rightarrow \cos \theta =\frac{2}{x^2}$     M1

 

Note:     This M1 may be seen anywhere, including a sketch of an appropriate triangle.

 

so $\frac{\theta }{4}+c=\frac{1}{4}\arccos \left(\frac{2}{x^2}\right)+c$     AG

[7 marks]

[N/A]